
A related version of the above theorem also exists. Define
\(\lbrace \psi _n\rbrace _{n\in \mathbb {N}}\subset F(K;X):=\lbrace \text{functions $\psi :K\rightarrow X$}\rbrace \) 
endowed with the uniform distance metric \(\sup _{t\in K}d_X(\cdot ,\cdot )\) . Assume that \(\lbrace \psi _n\rbrace _{n\in \mathbb {N}}\)  satisfies condition (i) and condition (ii) replaced by: for all \(\varepsilon >0\)  there exists \(\eta >0\)  such that
\(d_K(t,s)<\eta \quad \textup {\) 
0tsT\(}\quad \Rightarrow \quad \lim _{n\rightarrow \infty }d_X(\psi _n(t),\psi _n(s))< \varepsilon \quad \textup {\) nM\(}.\)\(Then there exists \) {nk}k\(\mathbb {N}\) F(K;X)\( such that \) nk\( in \) F(K;X)\( as \) k\(, and moreover, \) C(K;X)\(. We refer the reader to \cite [Theorem C.11]{DrEyGaGuHe18} for a proof.\item Note that by part (a) it is possible to conclude that the limit is indeed in \) C(K;X)\( even though the convergence is not in that space.\)
The standard Kolmogorov-Riesz's compactness theorem (see e.g. Theorem 5 in [1]}) is given as follows:
Theorem 1.3 (Kolmogorov-Riesz) 
Let \(p\in [1,\infty )\) . Then \(M\subset L^p(\mathbb {R}^N)\)  is relatively compact if and only if the following two conditions are fulfilled:

(Equicontinuity) For all \(\varepsilon >0\)  there is some \(\eta >0\)  such that, for all \(\psi \in M\)  and all \(\xi \in \mathbb {R}^N\)  with \(|\xi |<\eta \) ,
\(\Vert \psi (\cdot +\xi )-\psi \Vert _{L^p(\mathbb {R}^N)}< \varepsilon .\) 

(Equitightness) For all \(\varepsilon >0\)  there is some \(R>0\)  such that, for all \(\psi \in M\) ,
\(\int _{|x|> R}|\psi (x)|^p\,\mathrm {d}x<\varepsilon ^p.\) 


Remark 1.4 Recall that in [2]} it has been pointed out that the additional usual assumption of equiboundedness follows from properties (i)–(ii).

The Minkowski inequality for integrals
This is of course a well-known result (see e.g. Proposition 4.3 in [3]}), but we include it for completeness.
Theorem 2.1 
Let \((X,\mathcal {M},\mu )\)  and \((Y,\mathcal {N},\nu )\)  be \(\sigma \) -finite measure spaces and \(p\in [1,\infty ]\) .

If \(f\ge 0\)  is an \(\mathcal {M}\otimes \mathcal {N}\) -measurable function, then

the functions \(x\mapsto \Vert f(x,\cdot )\Vert _{L^1(Y,\nu )}\)  and \(y\mapsto \Vert f(\cdot ,y)\Vert _{L^p(X,\mu )}\)  are respectively \(\mu \) - and \(\nu \) -measurable; and


\(\begin{split}&\bigg \Vert \int _{Y}f(\cdot ,y)\,\mathrm {d}\nu (y)\bigg \Vert _{L^p(X,\mu )}=\bigg (\int _{X}\bigg |\int _{Y}f(x,y)\,\mathrm {d}\nu (y)\bigg |^p\,\mathrm {d}\mu (x)\bigg )^{\frac{1}{p}}\\&\le \int _{Y}\bigg (\int _{X}|f(x,y)|^p\,\mathrm {d}\mu (x)\bigg )^{\frac{1}{p}}\,\mathrm {d}\nu (y)=\int _{Y}\Vert f(\cdot ,y)\Vert _{L^p(x,\mu )}\,\mathrm {d}\nu (y).\end{split}\) 


If \(f:X\times Y\rightarrow \mathbb {C}\)  is an \(\mathcal {M}\otimes \mathcal {N}\) -measurable function and \(\int _{Y}\Vert f(\cdot ,y)\Vert _{L^p(X,\mu )}\,\mathrm {d}\nu (y)<\infty \) , then

the functions \(x\mapsto \Vert f(x,\cdot )\Vert _{L^1(Y,\nu )}\)  and \(y\mapsto \Vert f(\cdot ,y)\Vert _{L^p(X,\mu )}\)  are respectively \(\mu \) - and \(\nu \) -measurable;

for \(\mu \) -a.e. \(x\) , \(f(x,\cdot )\in L^1(Y,\,\mathrm {d}\nu )\) ;

the \(\mu \) -a.e. defined function \(x\mapsto \int _{Y}f(x,y)\,\mathrm {d}\nu (y)\)  is in \(L^p(X,\mu )\) ; and

the inequality in (a)(ii) holds.



(a) The desired measurability follows by Section 252P in [4]}, the rest is a consequence of the proof of Proposition 4.3 in [3]}.
(b) We apply part (a) to the function \(|f|\) .

Auxiliary results
Uniform estimates for discrete operators
In the local case \(\mathfrak {L}^{\sigma ,\mu }=\Delta \) , we can always control the discretization by the operator itself.
Lemma 3.1 
Assume \(p\in [1,\infty ]\) . Let \(\psi \in C^2(\mathbb {R}^N)\)  with \(\Delta \psi \in L^p(\mathbb {R}^N)\) , and recall that
\(\Delta _h\psi (x)=\sum _{i=1}^N\frac{\psi (x+he_i)+\psi (x-he_i)-2\psi (x)}{h^2}.\) 
Then
\(\begin{split}\Vert \Delta _h\psi \Vert _{L^p(\mathbb {R}^N)}\le 2\Vert \Delta \psi \Vert _{L^p(\mathbb {R}^N)}.\end{split}\) 

Remark 3.2 Here we immediately see that control of \(\Delta \psi \)  gives control of \(\Delta _h\psi \) .

An application of the Taylor theorem yields (\(|he_i|=h<1\) )
\(\begin{split}\Delta _h\psi (x)&=\sum _{i=1}^N\bigg (\int _{0}^1(1-t)D^2\psi (x+the_i)e_i\cdot e_i\,\mathrm {d}t+\int _{-1}^0(1+t)D^2\psi (x+the_i)e_i\cdot e_i\,\mathrm {d}t\bigg )\\&=\int _{-1}^1(1-|t|)\Delta \psi (x+the_i)\,\mathrm {d}t.\end{split}\) 
Finally, by Jensen's inequality and Tonelli's lemma,
\(\begin{split}\Vert \Delta _h\psi \Vert _{L^p(\mathbb {R}^N)}&=\bigg (\int _{\mathbb {R}^N}|\Delta _h\psi (x)|^p\bigg )^{\frac{1}{p}}\le \bigg (\int _{\mathbb {R}^N}\int _{-1}^1(1-|t|)^p|\Delta \psi (x+the_i)|^p\,\mathrm {d}t \,\mathrm {d}x\bigg )^{\frac{1}{p}}\\&\le 2\bigg (\int _{\mathbb {R}^N}|\Delta \psi (x)|^p \,\mathrm {d}x\bigg )^{\frac{1}{p}}=2\Vert \Delta \psi \Vert _{L^p(\mathbb {R}^N)}.\end{split}\) 
The proof is complete.
In the nonlocal case \(\mathfrak {L}^{\sigma ,\mu }=\mathcal {L}^\mu \) , we cannot argue as above, but we can give estimates on its discretizations in an analogous way.
Lemma 3.3 
Assume \(p\in [1,\infty ]\) . Let \(\psi \in C^2(\mathbb {R}^N)\)  with \(\psi , D\psi , D^2\psi \in L^p(\mathbb {R}^N)\) . Then:

Under \(\sum _{\gamma \ne 0}\omega _{\gamma ,h}<\infty \)  and for all \(R>1\) ,
\(\begin{split}&\Vert \mathcal {L}^h[\psi ]\Vert _{L^p(\mathbb {R}^N)}\\&\le \Vert D^2\psi \Vert _{L^p(\mathbb {R}^N)}\bigg \lbrace \sum _{h\le |z_\gamma |\le R}|z_\gamma |^2\omega _{\gamma ,h}\bigg \rbrace +2\Vert \psi \Vert _{L^p(\mathbb {R}^N)}\bigg \lbrace \sum _{|z_\gamma |>R}\omega _{\gamma ,h}\bigg \rbrace .\end{split}\) 

Under \(\sum _{\gamma \ne 0}\min \lbrace |z_\gamma |^2,|z_\gamma |\rbrace \omega _{\gamma ,h}<\infty \) ,
\(\begin{split}&\Vert \mathcal {L}^h[\psi ]\Vert _{L^p(\mathbb {R}^N)}\\&\le \Vert D^2\psi \Vert _{L^p(\mathbb {R}^N)}\bigg \lbrace \sum _{h\le |z_\gamma |\le 1}|z_\gamma |^2\omega _{\gamma ,h}\bigg \rbrace +\Vert D\psi \Vert _{L^p}\bigg \lbrace \sum _{|z_\gamma |>1}|z_\gamma |\omega _{\gamma ,h}\bigg \rbrace .\end{split}\) 


Remark 3.4 We see that the estimation of \(\Vert \mathcal {L}^h[\psi ]\Vert _{L^p}\)  only relies on information about \(\Vert \psi \Vert _{L^p}\) , \(\Vert D\psi \Vert _{L^p}\) , and \(\Vert D^2\psi \Vert _{L^p}\) . However, there is no immediate control of \(\mathcal {L}^h[\psi ]\)  through \(\mathcal {L}^\mu [\psi ]\) . Such control requires a deeper analysis, which we provide in Appendix REF  below.

(a) Recall that
\(\mathcal {L}^h[\psi ](x)=\sum _{\gamma \ne 0} (\psi (x+z_\gamma )-\psi (x)) \omega _{\gamma ,h}\) 
As usual, since \(\omega _{\gamma ,h}=\omega _{-\gamma ,h}\) , we use
\(\sum _{h\le |z_\gamma |\le R}z_\gamma \cdot D\psi (x)\mathbf {1}_{|z_\gamma |\le R}\omega _{\gamma ,h}=0\) 
to take advantage of the Taylor theorem:
\(\sum _{h\le |z_\gamma |\le R} (\psi (x+z_\gamma )-\psi (x)) \omega _{\gamma ,h}=\sum _{h\le |z_\gamma |\le R}\bigg (\int _0^1(1-t)D^2\psi (x+tz_\gamma )z_\gamma \cdot z_\gamma \,\mathrm {d}t\bigg )\omega _{\gamma ,h}\) 
We now apply a Minkowski type of inequality for integrals (cf. e.g. Theorem REF ) to get
\(\begin{split}\Vert \mathcal {L}^h[\psi ]\Vert _{L^p(\mathbb {R}^N)}^p&\le \int _{\mathbb {R}^N}\bigg |\sum _{h\le |z_\gamma |\le R}\Big (\psi (x+z_\gamma )-\psi (x)-z_\gamma \cdot D\psi (x)\mathbf {1}_{|z_\gamma |\le R}\Big )\omega _{\gamma ,h}\bigg |^p\,\mathrm {d}x\\&\quad +\int _{\mathbb {R}^N}\bigg |\sum _{|z_\gamma |>R}\big (\psi (x+z_\gamma )-\psi (x)\big )\omega _{\gamma ,h}\bigg |^p\,\mathrm {d}x\\&\le \Vert D^2\psi \Vert _{L^p(\mathbb {R}^N)}\bigg \lbrace \sum _{h\le |z_\gamma |\le R}|z_\gamma |^2\omega _{\gamma ,h}\bigg \rbrace +2\Vert \psi \Vert _{L^p(\mathbb {R}^N)}\bigg \lbrace \sum _{|z_\gamma |>R}\omega _{\gamma ,h}\bigg \rbrace .\end{split}\) 
(b) Here, we instead use the fundamental theorem of calculus when \(|z_\gamma |>1\) :
\(\psi (x+z_\gamma )-\psi (x)=\int _0^1D\psi (x+tz_\gamma )\cdot tz_\gamma \,\mathrm {d}t.\) 
Otherwise, the proof can then be completed in a similar manner as part (a).

Estimates for test function giving equitightness
An immediate consequences of Lemmas REF  and REF  is:
Corollary 3.5 
Assume (REF ), \(\mathcal {L}^h=\Delta _h\) , and let \(\mathcal {X}_R\)  be given by (REF )–(REF ).
If \(p>N/2\) , then
\(\sup _{0<h<1}\Vert \mathcal {L}^h[\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}\rightarrow 0\qquad \text{as $R\rightarrow \infty $}.\) 

The nonlocal case is more delicate.
Corollary 3.6 
Let \(\mathcal {X}_R\)  be given by (REF )–(REF ).
Then:

Under (REF ), \(p>N/2\) , and
\(\sup _{0<h<1}\bigg \lbrace R^{\frac{N}{p}-2}\sum _{1<|z_\gamma |\le R}|z_\gamma |^2\omega _{\gamma ,h}+R^{\frac{N}{p}}\sum _{|z_\gamma |>R}\omega _{\gamma ,h}\bigg \rbrace =o_R(1),\) 
we have
\(\sup _{0<h<1}\Vert \mathcal {L}^h[\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}\rightarrow 0\qquad \text{as $R\rightarrow \infty $}.\) 

Under (REF ) and \(p>N\) ,
\(\sup _{0<h<1}\Vert \mathcal {L}^h[\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}\rightarrow 0\qquad \text{as $R\rightarrow \infty $}.\) 


(a) Choose \(\psi =1-\mathcal {X}_R\)  in Lemma REF (a) and recall Lemma REF  to obtain
\(\begin{split}\sup _{0<h<1}\Vert \mathcal {L}^h[\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}&=\sup _{0<h<1}\Vert \mathcal {L}^h[1-\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}\\&\le R^{\frac{N}{p}-2}\Vert D^2\mathcal {X}\Vert _{L^p(\mathbb {R}^N)}\sup _{0<h<1}\bigg \lbrace \sum _{h\le |z_\gamma |\le 1}|z_\gamma |^2\omega _{\gamma ,h}\bigg \rbrace \\&\quad +\Big (\Vert D^2\mathcal {X}\Vert _{L^p(\mathbb {R}^N)}+2\Vert 1-\mathcal {X}\Vert _{L^p(\mathbb {R}^N)}\Big )\times \\&\quad \times \sup _{0<h<1}\bigg \lbrace R^{\frac{N}{p}-2}\sum _{1< |z_\gamma |\le R}|z_\gamma |^2\omega _{\gamma ,h}+R^{\frac{N}{p}}\sum _{|z_\gamma |>R}\omega _{\gamma ,h}\bigg \rbrace .\end{split}\) 
Both terms go to zero as \(R\rightarrow \infty \)  by \((\ref {Dmuas})\)  and \(p>N/2\) , and (REF ), respectively.
(b) Now, choose \(\psi =1-\mathcal {X}_R\)  in Lemma REF (b) and recall Lemma REF  to obtain
\(\begin{split}\sup _{0<h<1}\Vert \mathcal {L}^h[\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}&=\sup _{0<h<1}\Vert \mathcal {L}^h[1-\mathcal {X}_R]\Vert _{L^p(\mathbb {R}^N)}\\&\le \Big (R^{\frac{N}{p}-2}\Vert D^2\mathcal {X}\Vert _{L^p(\mathbb {R}^N)}+2R^{\frac{N}{p}-1}\Vert D\mathcal {X}\Vert _{L^p(\mathbb {R}^N)}\Big )\times \\&\quad \times \sup _{0<h<1}\bigg \lbrace \sum _{1< |z_\gamma |\le R}|z_\gamma |^2\omega _{\gamma ,h}+\sum _{|z_\gamma |>R}|z_\gamma |\omega _{\gamma ,h}\bigg \rbrace .\end{split}\) 
Since \(p>N=\max \lbrace N/2,N\rbrace \)  and (REF ) holds, the right-hand side goes to zero as \(R\rightarrow \infty \) .
Example 3.1 
To get some intuition on the assumption (REF ), let us think in terms of \(\mu \) , and check what happens for the choice
\(\,\mathrm {d}\mu _{\beta ,\alpha }(z):=\frac{\,\mathrm {d}z}{|z|^{N+\beta }}\mathbf {1}_{0<|z|\le 1}+\frac{\,\mathrm {d}z}{|z|^{N+\alpha }}\mathbf {1}_{|z|>1}\) 
with \(\beta ,\alpha \in (0,2)\) . As this measure satisfies (REF ), it is a Lévy measure, and \(\int _{0<|z|\le 1}|z|^{2}\,\mathrm {d}\mu _{\beta ,\alpha }(z)\)  is simply finite. Thus, the tail behaviour is more interesting to us. One can check that
\(\begin{split}R^{\frac{N}{p}-2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu _{\beta ,\alpha }(z)&=\frac{1}{2-\alpha }\big (R^{\frac{N}{p}-\alpha }-R^{\frac{N}{p}-2}\big )=\mathcal {O}(R^{\frac{N}{p}-\alpha })\\\end{split}\) 
and furthermore,
\(R^{\frac{N}{p}}\int _{|z|>R}\,\mathrm {d}\mu _{\beta ,\alpha }(z)=\mathcal {O}(R^{\frac{N}{p}-\alpha }).\) 
To ensure that \(\mathcal {L}^{\mu _{\beta ,\alpha }}[\mathcal {X}_R]\rightarrow 0\)  in \(L^p\)  as \(R\rightarrow \infty \) , we let \(p>N/\alpha >N/2\)  when \(N\ge 2\) , \(p>1/\alpha \)  when \(\alpha \in (0,1)\)  and \(N=1\) , and \(p>1\)  when \(\alpha \in [1,2)\)  and \(N=1\) .
Now, for any \(\mu \)  satisfying (REF ), we still have
\(R^{\frac{N}{p}-2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)=R^{\frac{N}{p}-\alpha }R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)\le CR^{\frac{N}{p}-\alpha }\) 
and
\(R^{\frac{N}{p}}\int _{|z|>R}\,\mathrm {d}\mu (z)=R^{\frac{N}{p}-\alpha }R^{\alpha }\int _{1<|z|\le R}\,\mathrm {d}\mu (z)\le CR^{\frac{N}{p}-\alpha }.\) 
Hence, the mentioned \(L^p\) -convergence still holds under this assumption as well.


Consistency and uniform estimates in the nonlocal case
In the local case, one can always relate the discrete operator with the limit operator \(\Delta \) , as in Lemma REF . This is not that clear in the nonlocal case. However, we will show that we can check (REF ), (REF ), and (REF ) through information about \(\mathcal {L}^\mu \) .
Proposition 3.2 in [6]} provides the following result: Assumption (REF ) holds if (REF ) holds and, for all \(\psi \in C_\textup {c}^\infty (\mathbb {R}^N)\) ,
\(\mathcal {L}^h[\psi ]\rightarrow \mathfrak {L}^{\sigma ,\mu }[\psi ] \qquad \textup {pointwise as h\rightarrow 0^+.}\) 
The proof is based on choosing the particular test function
\(\psi (x):={\left\lbrace \begin{array}{ll}|x|^2-1 \qquad \qquad &\text{if $|x|\le 1$}\\\phi (x) \qquad \qquad &\text{if $|x|> 1$,}\end{array}\right.}\) 
where \(\phi \)  is some nonnegative function such that \(\psi \in C_\textup {c}^\infty (\mathbb {R}^N)\) . We will now provide a similar result in the case (REF ):
Lemma 3.7 Assumption (REF ) holds if (REF ) holds and
for
\(\psi (x):={\left\lbrace \begin{array}{ll}|x|^2-1 \qquad \qquad &\text{if $|x|\le 1$}\\2(|x|-1) \qquad \qquad &\text{if $|x|> 1$,}\end{array}\right.}\) 
we have (REF ).

We follow the proof of Proposition 3.2 in [6]}. In our particular case,
\(D\psi (x)={\left\lbrace \begin{array}{ll}2x \qquad \qquad &\text{if $|x|\le 1$}\\2\frac{x}{|x|} \qquad \qquad &\text{if $|x|> 1$,}\end{array}\right.}\qquad \text{and}\qquad D^2\psi (x)={\left\lbrace \begin{array}{ll}2 \qquad \qquad &\text{if $|x|\le 1$}\\0 \qquad \qquad &\text{if $|x|> 1$.}\end{array}\right.}\) 
We get
\(\begin{split}\mathcal {L}^{\nu _h}[\psi ](0)&=\int _{|z|\le 1}\int _0^1(1-t)z^TD^2\psi (tz)z\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{|z|>1}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&=\int _{|z|\le 1}\int _0^1(1-t)z^TD^2\psi (tz)z\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&\quad +\int _{|z|>1/t}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&=\int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^12t^2|z|^2\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{|z|>1/t}\int _0^12t|z|\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&\ge \int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^12t^2\,\mathrm {d}t|z|\,\mathrm {d}\nu _h(z)+\int _{|z|>1/t}\int _0^12t\,\mathrm {d}t|z|\,\mathrm {d}\nu _h(z)\\&\ge \int _{|z|>0}\min \lbrace |z|^2,|z|\rbrace \,\mathrm {d}\nu _h(z).\end{split}\) 
Hence,
\(\begin{split}\sup _{0<h<1}\int _{|z|>0}\min \lbrace |z|^2,|z|\rbrace \,\mathrm {d}\nu _h(z)&\le \sup _{0<h<1}\mathcal {L}^{\nu _h}[\psi ](0)\\&\le \Vert \mathcal {L}^\mu [\psi ]\Vert _{L^\infty (\mathbb {R}^N)}+\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|,\end{split}\) 
which is finite due to the pointwise convergence and the assumption (REF ).
In the case (REF ), we get:
Lemma 3.8 Assumption (REF ) holds if \(\mu \)  satisfies (REF ) and for
\(\psi (x):={\left\lbrace \begin{array}{ll}R^{\alpha -2}|x|^2-R^\alpha \qquad \qquad &\text{if $|x|\le R$}\\\phi (x) \qquad \qquad &\text{if $|x|> R$,}\end{array}\right.}\) 
where \(\phi \)  is some nonnegative function such that \(\psi \in C_\textup {c}^\infty (\mathbb {R}^N)\)  and \(\Vert \psi \Vert _{L^\infty (\mathbb {R}^N)}=|\psi (0)|=R^\alpha \) , we have (REF ) uniformly in \(R\) .

Again, we follow the proof of Proposition 3.2 in [6]}. We obtain
\(\begin{split}\mathcal {L}^{\nu _h}[\psi ](0)&\ge R^{\alpha -2}\int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)-\psi (0)\int _{|z|>R}\,\mathrm {d}\nu _h(z)\\&\ge R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)+R^\alpha \int _{|z|>R}\,\mathrm {d}\nu _h(z).\end{split}\) 
Hence, since \(R>1\)  and \(\alpha \in (0,2)\) , we have \(R^{\alpha -2}\le 1\)  and
\(\begin{split}&\sup _{0<h<1}\bigg \lbrace R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\nu _h(z)+R^\alpha \int _{|z|>R}\,\mathrm {d}\nu _h(z)\bigg \rbrace \\&\le \sup _{0<h<1}\mathcal {L}^{\nu _h}[\psi ](0)\le |\mathcal {L}^\mu [\psi ](0)|+\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|\\&\le \bigg (R^{\alpha -2}\int _{|z|\le 1}|z|^2\,\mathrm {d}\mu (z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)+\int _{|z|>R}|\psi (z)-\psi (0)|\,\mathrm {d}\mu (z)\bigg )\\&\quad +\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|\\&\le \bigg (\int _{|z|\le 1}|z|^2\,\mathrm {d}\mu (z)+R^{\alpha -2}\int _{1<|z|\le R}|z|^2\,\mathrm {d}\mu (z)+2R^\alpha \int _{|z|>R}\,\mathrm {d}\mu (z)\bigg )\\&\quad +\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|,\end{split}\) 
which is finite due to the pointwise convergence and the assumption (REF ).
